Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $x = \dfrac{10n^2 + 100n}{-8n + 16} \div \dfrac{n^2 + 5n - 50}{n - 5} $
Dividing by an expression is the same as multiplying by its inverse. $x = \dfrac{10n^2 + 100n}{-8n + 16} \times \dfrac{n - 5}{n^2 + 5n - 50} $ First factor the quadratic. $x = \dfrac{10n^2 + 100n}{-8n + 16} \times \dfrac{n - 5}{(n + 10)(n - 5)} $ Then factor out any other terms. $x = \dfrac{10n(n + 10)}{-8(n - 2)} \times \dfrac{n - 5}{(n + 10)(n - 5)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ 10n(n + 10) \times (n - 5) } { -8(n - 2) \times (n + 10)(n - 5) } $ $x = \dfrac{ 10n(n + 10)(n - 5)}{ -8(n - 2)(n + 10)(n - 5)} $ Notice that $(n - 5)$ and $(n + 10)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ 10n\cancel{(n + 10)}(n - 5)}{ -8(n - 2)\cancel{(n + 10)}(n - 5)} $ We are dividing by $n + 10$ , so $n + 10 \neq 0$ Therefore, $n \neq -10$ $x = \dfrac{ 10n\cancel{(n + 10)}\cancel{(n - 5)}}{ -8(n - 2)\cancel{(n + 10)}\cancel{(n - 5)}} $ We are dividing by $n - 5$ , so $n - 5 \neq 0$ Therefore, $n \neq 5$ $x = \dfrac{10n}{-8(n - 2)} $ $x = \dfrac{-5n}{4(n - 2)} ; \space n \neq -10 ; \space n \neq 5 $